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\title{第二次作业}
\author{付临\\3200104960\\信息与计算科学}

\begin{document}
\maketitle
\section{}
解：\\
$\bullet$ 由线性插值可得$p_{1}(f;x)=-\frac{1}{2}x+\frac{3}{2}$，因为$f(x)=\frac{1}{x}$，则
$f''(\xi (x)) = \frac{2}{\xi^{3}(x)}$，则
\begin{equation*}
    \frac{1}{x}-(-\frac{1}{2}x+\frac{3}{2}) = \frac{1}{\xi^{3}{x}}(x-1)(x-2)
\end{equation*}
故得：$\xi(x) = (2x)^{\frac{1}{3}}$

$\bullet$ 因为在[1,2]内$\xi(x)$单调递增，则$\max \xi(x)=\sqrt[3]{4}$，$\min \xi(x)=\sqrt[3]{2}$。
又因为$f''(\xi(x))=\frac{1}{x}$，则$\max f''(\xi(x))=$1。

\section{}
解：\\
设$p_{n}(x)$为关于$x_{0},x_{1},\ldots ,x_{n}$的插值多项式。
\\令$p_{n}(x_{i})=\sqrt{f_{i}}$，且$p(x)=p_{n}^{2}(x)$，
\\则$p\in P_{2n}^{+}$，且$p_{x_{i}}=f_{i}$。

\section{}
解：\\
$\bullet$当$n=1$时，$f[t,t+1]=(e-1)e^{t}$，设当$n=k$时，等式成立，则当$n=k+1$时：
\begin{align*}
    f[t,t+1,\ldots,t+k+1]&=\frac{f[t+1,\ldots,t+k+1]-f[t,\ldots,t+k]}{k+1}\\
    &=\frac{\frac{(e-1)^{k}}{k!}e^{t+1}-\frac{(e-1)^{k}}{k!}e^{t}}{k+1}\\
    &=\frac{(e-1)^{k+1}}{(k+1)!}e^{t}
\end{align*}
则原式成立。


$\bullet$
\begin{align*}
    f[0,1,\ldots,n]&=\frac{(e-1)^{n}}{n!}e^{0}\\
    f[0,1,\ldots,n]&=\frac{1}{n!}f^{(n)}(\xi)\\
    (e-1)^{n}&=f^{(n)}(\xi)\\
    (e-1)^{n}&=e^{3}
\end{align*}
则$\xi=n\ln(e-1)>\frac{n}{2}$，即$\xi$在$\frac{n}{2}$的右侧。

\section{}
解：\\
$\bullet$由牛顿插值公式得：
\begin{equation*}
    \begin{array}{c|cccc}
    0 & 5 & & & \\
    1 & 3 & -2 & & \\
    3 & 5 & 1 & 1 & \\
    4 & 12 & 7 & 2 & \frac{1}{4}
    \end{array}
\end{equation*}
故$p_{3}(f;x)=5-2x+x(x-1)+\frac{1}{4}x(x-1)(x-3)=\frac{1}{4}x^{3}-\frac{9}{4}x+5$


$\bullet$令$p_{3}'(f;x)=\frac{3}{4}x^{2}-\frac{9}{4}>0$，得$x>\sqrt{3}$，所以
在$(1,\sqrt{3})$上$p_{3}\uparrow$，在$(\sqrt{3},3)$上$p_{3}\downarrow$，故$x_{min}=\sqrt{3}$。

\section{}
解：\\
$\bullet$因为$f(0)=0,f(1)=1,f(2)=128,f'(1)=7,f''(42),f'(2)=448$则：
\begin{equation*}
    \begin{array}{c|cccccc}
        0 & 0 & & & & & \\
        1 & 1 & 1 & & & & \\
        1 & 1 & 7 & 6 & & & \\
        1 & 1 & 7 & 21 & 15 & & \\
        2 & 128 & 127 & 120 & 99 & 42 & \\
        2 & 128 & 448 & 321 & 201 & 102 & 30
    \end{array}
\end{equation*}
故$f[0,1,1,1,2,2]=30$

$\bullet$由题意知：
\begin{align*}
    f[0,1,1,1,2,2]&=\frac{f^{(5)}(\xi)}{5!}\\
    f^{(5)}(\xi)=5!\times 30 &=7\times6\times5\times4\times3\times\xi^{2}\\
\end{align*} 
故：$\xi^{2}=\frac{5}{14},\xi=\frac{\sqrt{70}}{14}$

\section{}
解：\\
$\bullet$由题意可得：
\begin{equation*}
    \begin{array}{c|ccccc}
        0 & 1 & & & & \\
        1 & 2 & 1 & & & \\
        1 & 2 & -1 & -2 & & \\
        3 & 0 & -1 & 0 & \frac{2}{3} & \\
        3 & 0 & 0 & \frac{1}{2} & \frac{1}{4} & -\frac{5}{36}
    \end{array}
\end{equation*}
则$p_{4}(f;x)=1+x-2x(x-1)+\frac{2}{3}x(x-1)^{2}-\frac{5}{36}x(x-1)^{2}(x-3)$，
$f(2)\approx p_4(2)=\frac{11}{18}$

$\bullet$
\begin{align*}
    \because R_{4}(f;x)&=\frac{f^{(5)}(\xi)}{5!}(x-0)(x-1)^{2}(x-3)^{2}\\
    |f^{(5)}(x)|&\leq M\\
    \therefore |R_{4}(f;x)&\leq \frac{M}{5!}\times 2 = \frac{M}{60}
\end{align*}

\section{}
Proof:\\
$\bullet$
因为$\bigtriangleup f(x)=f(x+h)-f(x)=f(x_{1})-f(x_{0})=hf[x_{0},x_{1}]$\\
不妨设当$k$时，$\bigtriangleup ^{k}f(x)=k!h^{k}f[x_{0},\ldots,x_{k}]$\\
则当$k+1$时，有
\begin{align*}
    \bigtriangleup ^{k+1}f(x)&=\bigtriangleup \bigtriangleup ^{k}f(x)\\
    &=\bigtriangleup ^{k}f(x+h)-\bigtriangleup ^{k}f(x)\\
    &=k!h^{k}f[x_{1},\ldots,x_{k+1}]-k!h^{k}f[x_{0},\ldots,x_{k}]\\
    &=(k+1)!h^{k+1}f[x_{0},\ldots,x_{k+1}]
\end{align*}
即可证$\bigtriangleup ^{k}f(x)=k!h^{k}f[x_{0},\ldots,x_{k}]$\\
$\bullet$ 
因为$\bigtriangledown f(x)=f(x)-f(x-h)-f(x_{0})-f(x_{-1})=hf[x_{0},x_{-1}]$\\
不妨设当$k$时，$\bigtriangledown ^{k}f(x)=k!h^{k}f[x_{0},\ldots,x_{-k}]$\\
则当$k+1$时，有
\begin{align*}
    \bigtriangledown ^{k+1}f(x)&=\bigtriangledown \bigtriangledown ^{k}f(x)\\
    &=\bigtriangledown ^{k}f(x)-\bigtriangledown ^{k}f(x-h)\\
    &=k!h^{k}f[x_0,\ldots,x_{-k}]-k!h^{k}f[x_{-1},\ldots,x_{-k-1}]\\
    &=(k+1)!h^{k+1}f[x_{0},\ldots,x_{-k-1}]
\end{align*}
即可证$\bigtriangledown ^{k}f(x)=k!h^{k}f[x_{0},\ldots,x_{-k}]$

\section{}
Proof:\\
\begin{align*}
    \frac{\partial}{\partial x_{0}}f[x_{0},\ldots,x_{n}]&=\lim_{h \to 0} \frac{f[x_{0}+h,\ldots,x_{n}]-f[x_{0},\ldots,x_{n}]}{h}\\
    &=\lim_{h \to 0} \frac{f[x_{1},\ldots,x_{n},x_{0}+h]-f[x_{0},\ldots,x_{n}]}{h}\\
    &=\lim_{h \to 0} f[x_{0},\ldots,x_{n},x_{0}+h]\\
    &=f[x_{0},x_{0},x_{1},\ldots,x_{n}]
\end{align*}
对$\forall k=0,1,\ldots,n$，若$f$在$x_{k}$处可微，则
\begin{align*}
    \frac{\partial}{\partial x_{k}}f[x_{0},\ldots,x_{n}]=f[x_{0},x_{1},\ldots,x_{k},x_{k},x_{k+1},\ldots,x_{n}]
\end{align*}   

\section{}
解：\\
因为$\max_{x\in[-1,1]} |x^{n}+a_{1}x^{n-1}+\ldots+a_{n}|\geq \frac{1}{2^{n-1}}$，
则对$x$做线性变换，令$x=\frac{b-a}{x}t+\frac{a+b}{2}$，因为$x\in[a,b]$，则$t\in[-1,1]$，故
\begin{align*}
    \min \max_{x\in [a,b]} |a_{0}x^{n}+\ldots+a_{n}|&=\min \max_{t\in [-1,1]} |a_{0}(\frac{b-a}{x})^{n}t^{n}+\ldots+a_{n}'|\\
    &=\min \max_{t \in [-1,1]} |a_{0}(\frac{b-a}{2}^{n})||t^{n}+\ldots+a_{n}''|\\
    &=\frac{|a_{0}(\frac{b-a}{2})^{n}|}{2^{n-1}}
\end{align*}

\section{}
Proof:\\
因为$T_{n}(a)$的$a^{n}$次项系数为$2^{n-1}$，且$a>1$，则
\begin{align*}
    |\frac{T_{n}(x)}{T_{n}(a)}|=\frac{|T_{n}(x)|}{|T_{n}(a)|}<\frac{|T_{n}(x)}{2^{n-1}}\\
    \because \max_{x\in [-1,1]} |\frac{T_n{x}}{2^{n-1}}|\leq \max_{x\in [-1,-1]}|p(x)|\\
    \therefore  \forall p \in \mathbb{P}_n^a \qquad \|\hat{p}_n\|_{\infty} \leq \|p\|_{\infty}
\end{align*}

\section{}
Proof:\\
$\bullet \because b_{n,k}(t)=C_{n}^{k}t^{k}(1-k)^{n-k} \qquad t\in [0,1]\\
\therefore b_{n,k}(t)>0\\
\bullet \sum_{k = 1}^{n} b_{n,k}(t)=\sum_{k=0}^{n}C_{n}^{k}t^{k}(1-t)^{n-k}=(t+1-t)^{n}=1\\  
\bullet$
\begin{align*}
    \sum_{k=0}^{n}kb_{n,k}(t)&=\sum_{k=0}^{n}kC_{n}^{k}t^{k}(1-t)^{n-k}\\
    &=\sum_{k=1}^{n}ntC_{n-1}^{k-1}t^{k-1}(1-t)^{n-k}\\
    &=nt(t+1-t)^{n-1}\\
    &=nt
\end{align*}\\
$\bullet$
\begin{align*}
    \sum_{k=0}^{n}(k-nt)^{2}b_{n,k}(t)&=\sum_{k=0}^{n}(k-nt)^{2}C_{n}^{k}t^{k}(1-t)^{n-k}\\
    &=\sum_{k=0}^{n}(k^{2}-2knt+n^{2}t^{2})C_{n}^{k}t^{k}(1-t)^{n-k}\\
    \sum_{k=0}^{n}k^{2}b_{n,k}(t)&=nt\sum_{k=1}^{n}kb_{n-1,k-1}(t)\\
    &=nt\sum_{k=1}^{n}[(k-1)b_{n-1,k-1}(t)+b_{n-1,k-1}(t)]\\
    &=nt[(n-1)t+1]\\
    \therefore \sum_{k=0}^{n}(k-nt)^{2}b_{n,k}(t)&=nt[(n-1)t+1]-2ntnt+n^{2}t^{2}\\
    &=nt(1-t)
\end{align*}
\end{document}
